Question

Помогите срочно надо пожалуйста ​

Answer 1

1.

$\int\limits^{ 1 } _ {0}(2 {x}^{3} + 4x)dx = ( \frac{2 {x}^{4} }{4} + \frac{4 {x}^{2} }{2} )| ^{1} _ {0} = \\ = ( \frac{ {x}^{4} }{2} + 2 {x}^{2}) | ^{1} _ {0} = \frac{1}{2} + 2 - 0 = 2.5$

2.

$\int\limits^{ 1 } _ {0} \frac{dx}{ \sqrt{4 + 3 {x}^{2} } } = \int\limits^{ 1} _ {0} \frac{dx}{ {( \sqrt{3} x)}^{2} + {2}^{2} } = \\ = \frac{1}{ \sqrt{3} } \int\limits^{ 1} _ {0} \frac{d( \sqrt{3} x)}{ \sqrt{ {( \sqrt{3}x) }^{2} + {2}^{2} } } = \frac{1}{ \sqrt{3} } ln( \sqrt{3}x + \sqrt{3{x}^{2} + 4} ) | ^{1} _ {0} = \\ = \frac{1}{ \sqrt{3} }( ln( \sqrt{3} + \sqrt{7} ) - ln(0 + 2) ) = \\ = \frac{1}{ \sqrt{3} } ln( \frac{ \sqrt{3} + \sqrt{7} }{2} )$

3.

$\int\limits^{ 2 } _ {1} {( {x}^{2} - 1)}^{3} xdx = \frac{1}{2} \int\limits^{ 2 } _ {1} {( {x}^{2} - 1)}^{3}2xdx = \\ = \frac{1}{2} \int\limits^{ 2} _ {1} {( {x}^{2} - 1) }^{3} d( {x}^{2} - 1) = \frac{ {( {x}^{2} - 1)}^{4} }{8} | ^{2} _ {1} = \\ = \frac{1}{8} ( {(4 - 1)}^{4} - 0) = \frac{1}{8} \times 81 = \frac{81}{8}$

4.

$\int\limits^{ \frac{\pi}{2} } _ {0} \frac{ \cos(x) }{4 - \sin(x) }dx = - \int\limits^{ \frac{\pi}{2} } _ {0} \frac{( - \cos(x)) }{ 4 - \sin(x) }dx = \\ = - \int\limits^{ \frac{\pi}{2} } _ {0} \frac{d(4 - \sin(x)) }{4 - \sin(x) } = - ln(4 - \sin(x) ) | ^{ \frac{\pi}{2} } _ {0} = \\ = - ln(4 - 1) + ln(4) = ln( \frac{4}{3} )$

5.

Решим неопределённый интеграл по частям:

$\int\limits \: x \cos(3x) dx \\ \\ u = x \: \: \: \: \: \: \: \: \: du = dx \\ dv = \cos(3x) dx \: \: \: \: \: v = \frac{1}{3} \int\limits \cos(3x) d(3x) = \\ = \frac{1}{3} \sin(3x) \\ \\ uv - \int\limits \: vdu = \\ = \frac{x}{3} \sin(3x) - \frac{1}{3} \int\limits \sin(3x) dx = \\ = \frac{x}{3} \sin(3x) + \frac{1}{9} \cos(3x) + c \\ \\ (\frac{x}{3} \sin(3x) + \frac{1}{9} \cos(3x) )| ^{ \frac{\pi}{4} } _ {0} = \\ = \frac{\pi}{12} \sin( \frac{3\pi}{4} ) + \frac{1}{9} \cos( \frac{3\pi}{4} ) - 0 - \frac{1}{9} = \\ = \frac{\pi}{12} \times \frac{ \sqrt{2} }{2} - \frac{1}{9} \times \frac{ \sqrt{2} }{2} - \frac{1}{9} = \\ = \frac{\pi \sqrt{2} }{2} - \frac{ \sqrt{2} }{9}$